CHAPTER 7 – TRIANGLES (Class 9, NCERT/CBSE)
1-Mark (20)
2-Mark (20)
3-Mark (20)
Exercise 7.1 – Solutions
Exercise 7.2 – Solutions
Exercise 7.3 – Solutions
Part A — 1-Mark Questions (20) with Solutions
Q1. What do you mean by congruent figures?
Two figures are congruent if they have the same shape and the same size; under a suitable correspondence all their corresponding parts are equal (CPCT).
Q2. State the SAS congruence rule.
If two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the triangles are congruent (SAS).
Q3. State the ASA congruence rule.
If two angles and the included side of one triangle are equal to two angles and the included side of another triangle, the triangles are congruent (ASA).
Q4. State the AAS congruence rule.
If two angles and one corresponding side of one triangle are equal to two angles and the corresponding side of another triangle, then the triangles are congruent (AAS).
Q5. State the SSS congruence rule.
If three sides of one triangle are equal to three sides of another triangle, the triangles are congruent (SSS).
Q6. State the RHS congruence rule.
In two right-angled triangles, if the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other, the triangles are congruent (RHS).
Q7. What is an isosceles triangle?
A triangle in which at least two sides are equal; usually \(AB=AC\) implies base \(BC\) and base angles at \(B\) and \(C\) are equal.
Q8. In an isosceles triangle with \(AB=AC\), which angles are equal?
Angles opposite equal sides: hence \(\angle B=\angle C\).
Q9. What are CPCT (or CPCTC)?
Corresponding Parts of Congruent Triangles are Congruent (Equal).
Q10. Can equality of three angles alone ensure triangle congruence?
No. Equal triangles by AAA are only similar, not necessarily congruent (side lengths can scale).
Q11. Each angle of an equilateral triangle equals?
\(60^\circ\).
Q12. If \(\triangle PQR \cong \triangle ABC\) with correspondence \(P\leftrightarrow A, Q\leftrightarrow B, R\leftrightarrow C\), then \(PR\) equals?
\(PR=AC\) (corresponding sides).
Q13. What is the included angle between sides \(AB\) and \(AC\) in \(\triangle ABC\)?
\(\angle A\).
Q14. Name the side opposite \(\angle A\) in \(\triangle ABC\).
Side \(BC\).
Q15. If two lines are each perpendicular to the same line, what can you say about them?
They are parallel.
Q16. Vertically opposite angles are always…?
Equal.
Q17. What is the hypotenuse in a right triangle?
The side opposite the right angle; the longest side.
Q18. Angle sum property of a triangle is?
\(\angle A+\angle B+\angle C=180^\circ\).
Q19. If \(AB=AC\) and \(AD\) is the bisector of \(\angle A\), what is the relation between \(BD\) and \(DC\)?
\(BD=DC\) (in an isosceles triangle, angle bisector from the vertex also bisects the base).
Q20. Which congruence rule is not valid: SSA or ASS?
Neither SSA nor ASS is a valid general congruence rule.
Part B — 2-Mark Questions (20) with Solutions
Q1. In \(\triangle ABC\), \(AB=AC\). Prove that the base angles are equal.
Draw \(AD\) the bisector of \(\angle A\) meeting \(BC\) at \(D\). In \(\triangle ABD\) and \(\triangle ACD\): \(AB=AC\) (given), \(\angle BAD=\angle CAD\) (construction), \(AD=AD\) (common). Hence triangles are congruent (SAS) and \(\angle B=\angle C\) (CPCT).
Q2. If two angles of a triangle are equal, prove the opposite sides are equal.
Let \(\angle B=\angle C\). Draw \(AD\) as bisector of \(\angle A\) meeting \(BC\) at \(D\). In \(\triangle ABD\) and \(\triangle ACD\): \(\angle BAD=\angle CAD\) (construction), \(\angle B=\angle C\) (given), \(AD\) common. By ASA, \(\triangle ABD\cong \triangle ACD\). Hence \(AB=AC\) (CPCT).
Q3. In \(\triangle ABC\), \(AD\) is perpendicular bisector of \(BC\). Show \(AB=AC\).
Given \(BD=DC\) and \(\angle ADB=\angle ADC=90^\circ\). In \(\triangle ADB\) and \(\triangle ADC\): \(BD=DC\) (given), \(AD\) common, right angle each; by RHS, triangles are congruent \(\Rightarrow AB=AC\).
Q4. Prove that the perpendicular bisector of a line segment is the locus of points equidistant from its endpoints.
Let \(P\) lie on perpendicular bisector of \(AB\). Then \(PA=PB\) (RHS on \(\triangle PAB\) and \(\triangle PBA\) with \(AD\perp AB\) at midpoint). Conversely, if \(PA=PB\), \(P\) lies on the perpendicular bisector of \(AB\) (construction and congruence as in text).
Q5. In right triangles \(ABC\) and \(PQR\) with \(\angle C=\angle R=90^\circ\), if \(AB=PQ\) and \(BC=QR\), show \(\triangle ABC\cong \triangle PQR\).
Given one leg and hypotenuse equal \( (\text{or } \)hypotenuse and one side\()\Rightarrow\) RHS congruence \(\Rightarrow\) triangles congruent.
Q6. In \(\triangle ABC\), if \(AB=AC\) and \(BE\perp AC\), \(CF\perp AB\). Prove \(BE=CF\).
Right triangles \(\triangle ABE\) and \(\triangle ACF\): \(AB=AC\) (given), right angles at \(E\) and \(F\), and \(A\) is common vertex with \(AE\) and \(AF\) as corresponding. By RHS, \(\triangle ABE\cong \triangle ACF\Rightarrow BE=CF\).
Q7. In \(\triangle ABC\), \(AB=AC\) and \(AD\) is median on \(BC\). Show \(AD\perp BC\).
In \(\triangle ABD\) and \(\triangle ACD\): \(AB=AC\) (given), \(BD=DC\) (median), \(AD\) common. By SSS, triangles congruent. Then \(\angle ADB=\angle ADC\) and they are a linear pair \(\Rightarrow \angle ADB=\angle ADC=90^\circ\), hence \(AD\perp BC\).
Q8. If two circles have equal radii, are they congruent? Justify.
Yes. All corresponding chords/arcs subtend equal angles and the radii match; a circle is determined by its radius, hence equal radii \(\Rightarrow\) congruent circles.
Q9. In \(\triangle ABC\), \(AB=AC\). If \(AP\) is angle bisector of \(\angle A\), prove \(BP=PC\).
As in isosceles triangle, \(AP\) is also perpendicular and median to base. Using congruence of \(\triangle ABP\) and \(\triangle ACP\) (SAS), get \(BP=PC\).
Q10. \(\triangle ABC\) with \(AB=AC\). If \(D\) and \(E\) are midpoints of \(AB\) and \(AC\), prove \(DE\parallel BC\).
By Midpoint Theorem (or using similar triangles from congruent halves), the segment joining midpoints of two sides is parallel to the third side: \(DE\parallel BC\).
Q11. If \(AD\) and \(BC\) are equal perpendiculars to segment \(AB\), prove that \(CD\) bisects \(AB\).
Right triangles \(\triangle ACD\) and \(\triangle BCD\) have \(AD=BC\) and a common hypotenuse \(CD\). By RHS, they are congruent; hence \(AC=CB\Rightarrow C\) is midpoint of \(AB\).
Q12. In \(\triangle ABC\), \(AB=AC\). Find \(\angle B\) and \(\angle C\) if \(\angle A=40^\circ\).
\(\angle B=\angle C\). Sum \(=180^\circ\Rightarrow \angle B=\angle C=\dfrac{180^\circ-40^\circ}{2}=70^\circ\).
Q13. \(\triangle ABC\) right-angled at \(C\) and \(M\) is midpoint of \(AB\). Show \(CM=\dfrac{AB}{2}\).
Right triangle median to hypotenuse equals half the hypotenuse. Using congruence of \(\triangle AMC\) and \(\triangle BMC\) (RHS), \(CM=AM=BM=\frac{AB}{2}\).
Q14. If two angles of one triangle equal two angles of another, prove that the third angles are equal.
Let \(\angle A=\angle P\) and \(\angle B=\angle Q\). Then \(\angle C=180^\circ-(\angle A+\angle B)=180^\circ-(\angle P+\angle Q)=\angle R\).
Q15. Show that the angle bisectors in an isosceles triangle are also medians to the base.
For \(AB=AC\), angle bisector \(AD\) gives congruent \(\triangle ABD\) and \(\triangle ACD\) (ASA/SAS). Hence \(BD=DC\); so \(AD\) is a median.
Q16. In \(\triangle ABC\), if \(AB=AC\) and \(E\) on \(AB\), \(F\) on \(AC\) with \(BE\parallel CF\), prove \(AE=AF\).
Corresponding angles equal \(\Rightarrow\) triangles \(AEB\) and \(AFC\) are similar. Since \(AB=AC\), the similarity ratio gives \(AE=AF\).
Q17. If \(AP\) is a common side of triangles \(APB\) and \(APC\), with \(PB=PC\) and \(\angle APB=\angle APC\), prove \(AB=AC\).
By SAS: \(PB=PC\) (given), included angle equal, and \(AP\) common \(\Rightarrow \triangle APB\cong \triangle APC\Rightarrow AB=AC\).
Q18. In \(\triangle ABC\), external point \(P\) satisfies \(PA=PB=PC\). Show that \(P\) lies on perpendicular bisectors of all sides.
From \(PA=PB\), \(P\) is on perpendicular bisector of \(AB\); similarly for other pairs. Intersection means circumcenter.
Q19. If \(\triangle ABC \cong \triangle PQR\) (SSS), write a correct correspondence.
Match longest to longest, etc. Example: if \(AB=PQ\), \(BC=QR\), \(CA=RP\), then \(A\leftrightarrow P\), \(B\leftrightarrow Q\), \(C\leftrightarrow R\); write \(\triangle ABC\cong \triangle PQR\).
Q20. In \(\triangle ABC\), \(AB\neq AC\). Can the bisector of \(\angle A\) be perpendicular to \(BC\)?
Only if \(AB=AC\). If angle bisector is perpendicular to base, congruent halves force \(AB=AC\). Hence for \(AB\neq AC\), it cannot be perpendicular.
Part C — 3-Mark Questions (20) with Solutions
Q1. In Fig., \(OA=OB\) and \(OD=OC\). Prove \(\triangle AOD\cong \triangle BOC\) and hence \(AD\parallel BC\).
In \(\triangle AOD\) and \(\triangle BOC\): \(OA=OB\) (given), \(OD=OC\) (given), \(\angle AOD=\angle BOC\) (vertically opposite). By SAS, triangles are congruent \(\Rightarrow \angle OAD=\angle OBC\) (CPCT) are alternate interior angles, hence \(AD\parallel BC\).
Q2. Line \(l\) is the perpendicular bisector of \(AB\). For any \(P\) on \(l\), prove \(PA=PB\).
Let \(l\) meet \(AB\) at \(M\) with \(AM=MB\) and \(l\perp AB\). In \(\triangle PAM\) and \(\triangle PBM\): \(AM=MB\), \(PM\) common, right angles at \(M\). Using RHS, \(\triangle PAM\cong \triangle PBM\Rightarrow PA=PB\).
Q3. In \(\triangle ABC\), \(AD\) and \(BE\) are medians. If \(AB=AC\), prove \(DE\parallel BC\).
From \(AB=AC\), median \(AD\) is also perpendicular bisector and angle bisector. Midpoint \(E\) of \(AC\) and midpoint \(D\) of \(BC\) imply by Midpoint Theorem that segment joining midpoints of two sides is parallel to the third side \(\Rightarrow DE\parallel BC\).
Q4. If \(l\parallel m\) and \(p\parallel q\), prove \(\triangle ABC\cong \triangle CDA\) as in Fig. (text).
Using alternate interior angles equality from parallel pairs, find two angle equalities and one included side equality \(CA\) common. Apply ASA to get \(\triangle ABC\cong \triangle CDA\).
Q5. In isosceles \(\triangle ABC\) (\(AB=AC\)), with \(E, F\) midpoints of \(AB, AC\), show \(BF=CE\).
Consider \(\triangle ABF\) and \(\triangle ACE\): \(AB=AC\) (given), \(AF=AE\) (midpoints), \(\angle A\) common. By SAS, triangles congruent \(\Rightarrow BF=CE\) (CPCT).
Q6. In isosceles \(\triangle ABC\) with \(AB=AC\), points \(D,E\) on \(BC\) satisfy \(BE=CD\). Prove \(AD=AE\).
Use SAS on \(\triangle ABD\) and \(\triangle ACE\): \(AB=AC\), included angles \(\angle B=\angle C\) (base angles equal), and \(BD=CE\) (since \(BE=CD\Rightarrow BD=CE\)). Hence congruent \(\Rightarrow AD=AE\).
Q7. In right triangle \(ABC\) (\(\angle C=90^\circ\)), \(M\) is midpoint of hypotenuse \(AB\). If \(CM\) is produced to \(D\) with \(DM=CM\), prove: (i) \(\triangle AMC\cong \triangle BMD\); (ii) \(\angle DBC=90^\circ\); (iii) \(\triangle DBC\cong \triangle ACB\); (iv) \(CM=\dfrac{AB}{2}\).
(i) \(AM=BM\) (midpoint), \(CM=DM\) (given), and \(AM\parallel DM\) in kite-like setting with sharing equal sides leads to SSS on \(\triangle AMC\) and \(\triangle BMD\).
(ii) From (i), corresponding angles at \(B\) and \(C\) give a right angle at \(B\).
(iii) With \(\angle DBC=90^\circ\) and common leg \(BC\), and \(BD=AC\) from (i), apply RHS to get \(\triangle DBC\cong \triangle ACB\).
(iv) In a right triangle, median to hypotenuse equals half the hypotenuse, hence \(CM=\dfrac{AB}{2}\).
(ii) From (i), corresponding angles at \(B\) and \(C\) give a right angle at \(B\).
(iii) With \(\angle DBC=90^\circ\) and common leg \(BC\), and \(BD=AC\) from (i), apply RHS to get \(\triangle DBC\cong \triangle ACB\).
(iv) In a right triangle, median to hypotenuse equals half the hypotenuse, hence \(CM=\dfrac{AB}{2}\).
Q8. In \(\triangle ABC\) and \(\triangle PQR\), \(AB=PQ\), \(BC=QR\), and medians \(AM=PN\) are equal. Prove \(\triangle ABC\cong \triangle PQR\).
First show \(\triangle ABM\cong \triangle PQN\) (SSS/ SAS using median halves). Then use CPCT to deduce \(AC=PR\). Now SSS on whole triangles gives \(\triangle ABC\cong \triangle PQR\).
Q9. From an external point \(P\), tangents \(PA\) and \(PB\) are drawn to a circle with center \(O\). Prove \(PA=PB\).
Right triangles \( \triangle POA\) and \( \triangle POB\) have \(OA=OB\) (radii), \(OP\) common, and right angles at \(A,B\). By RHS, congruent \(\Rightarrow PA=PB\).
Q10. In \(\triangle ABC\), if \(\angle B=\angle C\) and \(BD\) is perpendicular to \(AC\), prove \(CD=AD\).
Right triangles \(\triangle ABD\) and \(\triangle ACD\): \(\angle B=\angle C\) \(\Rightarrow\) acute angles at \(D\) also equal; with \(AD\) common, by AAS/ RHS-style reasoning we get congruence and hence \(CD=AD\).
Q11. If two sides and a non-included angle of one triangle equal the corresponding two sides and non-included angle of another, are the triangles always congruent?
No. SSA/ASS is not a valid general rule; ambiguous case exists (two possible triangles in general).
Q12. In \(\triangle ABC\), \(AB=AC\). If exterior angle at \(A\) equals \(x^\circ\), find \(\angle B\) and \(\angle C\).
Exterior angle at \(A\) equals sum of remote interior: \(x=\angle B+\angle C=2\angle B\Rightarrow \angle B=\angle C=\dfrac{x}{2}\).
Q13. Prove that the perpendicular drawn from the vertex of an isosceles triangle to the base bisects the vertex angle.
In isosceles \(\triangle ABC\) with \(AB=AC\), let \(AD\perp BC\). Congruence of \(\triangle ABD\) and \(\triangle ACD\) (RHS or SSS) gives \(\angle BAD=\angle CAD\), hence the vertex angle is bisected.
Q14. If \(\triangle ABC\cong \triangle DEF\) and \(AB=DE\), \(BC=EF\), \(AC=DF\), show corresponding angles are equal.
By SSS congruence, triangles are congruent \(\Rightarrow\) CPCT: \(\angle A=\angle D\), \(\angle B=\angle E\), \(\angle C=\angle F\).
Q15. In \(\triangle ABC\), \(AD\) bisects \(\angle A\) and meets \(BC\) at \(D\). If \(AB\neq AC\), compare \(BD\) and \(DC\).
By Angle Bisector Theorem, \(\dfrac{BD}{DC}=\dfrac{AB}{AC}\). If \(AB>AC\Rightarrow BD>DC\); if \(AB
Q16. \(\triangle ABC\) is right-angled at \(C\). Show that the midpoint of hypotenuse is equidistant from all three vertices.
Let \(M\) midpoint of \(AB\). By RHS congruence on \(\triangle AMC\) and \(\triangle BMC\), \(AM=BM=CM\). Thus \(M\) is circumcenter of the triangle.
Q17. Prove: if a transversal makes equal alternate interior angles with two lines, the lines are parallel.
Take triangles formed by the transversal. Equality of a pair of angles and a common side leads to ASA congruence; corresponding angles then equal \(\Rightarrow\) lines are parallel.
Q18. In \(\triangle ABC\), \(E\) and \(F\) are points on \(AC\) and \(AB\) such that \(BE\perp AC\) and \(CF\perp AB\). If \(BE=CF\), prove \(AB=AC\).
Right triangles \(\triangle ABE\) and \(\triangle ACF\) have equal altitudes \(BE=CF\) and right angles. If additionally \(\angle A\) is common, use RHS to deduce \(AB=AC\).
Q19. In \(\triangle ABC\), if \(AB=AC\) and \(BD\) and \(CE\) are medians, prove \(BD=CE\).
From congruent halves \(\triangle ABD\cong \triangle ACD\) (SSS/SAS), corresponding medians to equal sides are equal; or use symmetry about the altitude through \(A\).
Q20. In isosceles \(\triangle ABC\) with \(AB=AC\), the external bisector of \(\angle A\) meets \(BC\) at \(P\). Show \(BP=PC\).
Exterior angle bisector theorem gives \(\dfrac{BP}{PC}=\dfrac{AB}{AC}=1\Rightarrow BP=PC\).
Exercise 7.1 — Textbook Questions & Perfect Solutions
| Question | Solution |
|---|---|
| Q1. In quadrilateral \(ACBD\), \(AC=AD\) and \(AB\) bisects \(\angle A\). Show \(\triangle ABC\cong \triangle ABD\). What about \(BC\) and \(BD\)? | In \(\triangle ABC\) and \(\triangle ABD\): \(AC=AD\) (given), \(\angle CAB=\angle DAB\) (bisected by \(AB\)), \(AB\) common. By SAS, triangles congruent. Hence \(BC=BD\) (CPCT). |
| Q2. In quadrilateral \(ABCD\), \(AD=BC\) and \(\angle DAB=\angle CBA\). Prove (i) \(\triangle ABD\cong \triangle BAC\); (ii) \(BD=AC\); (iii) \(\angle ABD=\angle BAC\). |
(i) In \(\triangle ABD\) and \(\triangle BAC\): \(AD=BC\) (given), \(\angle DAB=\angle CBA\) (given), \(AB\) common \(\Rightarrow\) SAS \(\Rightarrow\) congruent. (ii) \(BD=AC\) (CPCT). (iii) \(\angle ABD=\angle BAC\) (CPCT). |
| Q3. \(AD\) and \(BC\) are equal perpendiculars to line segment \(AB\). Show that \(CD\) bisects \(AB\). | Consider \(\triangle ACD\) and \(\triangle BCD\): \(AD=BC\) (given), \(\angle ADC=\angle BCD=90^\circ\), \(CD\) common. RHS \(\Rightarrow\) congruent \(\Rightarrow AC=CB\). Thus \(C\) is midpoint of \(AB\); so \(CD\) is the perpendicular bisector (already perpendicular). |
| Q4. Lines \(l\parallel m\) are intersected by another pair \(p\parallel q\). Show \(\triangle ABC\cong \triangle CDA\) (figure as in text). | From parallelism: \(\angle ABC=\angle CDA\) (alternate interior), \(\angle ACB=\angle CAD\) (alternate interior), and \(AC\) common. By ASA, \(\triangle ABC\cong \triangle CDA\). |
| Q5. Line \(l\) bisects \(\angle A\); \(B\) on \(l\). \(BP\) and \(BQ\) are perpendiculars to the arms of \(\angle A\). Show (i) \(\triangle APB\cong \triangle AQB\); (ii) \(BP=BQ\). |
(i) \(\angle PBA=\angle ABQ=90^\circ\), \(\angle PAB=\angle QAB\) (since \(AB\) bisects \(\angle A\)), \(AB\) common. By AAS/RHS, congruent. (ii) \(BP=BQ\) (CPCT): point \(B\) is equidistant from arms of \(\angle A\). |
Exercise 7.2 — Textbook Questions & Perfect Solutions
| Question | Solution |
|---|---|
| Q1. In isosceles \(\triangle ABC\) with \(AB=AC\), bisectors of \(\angle B\) and \(\angle C\) meet at \(O\). Join \(AO\). Show (i) \(OB=OC\); (ii) \(AO\) bisects \(\angle A\). |
(i) \(O\) lies on internal bisectors of \(\angle B,\angle C\); in an isosceles triangle these bisectors are symmetric \(\Rightarrow OB=OC\) (equal distances from sides). (ii) Consider \(\triangle ABO\) and \(\triangle ACO\): \(AB=AC\) (given), \(\angle ABO=\angle OCA\) (each half of base angles), \(AO\) common. By SAS, congruent \(\Rightarrow \angle BAO=\angle OAC\), so \(AO\) bisects \(\angle A\). |
| Q2. In \(\triangle ABC\), \(AD\) is perpendicular bisector of \(BC\). Prove \(AB=AC\). | \(\triangle ABD\) and \(\triangle ACD\): \(BD=DC\) and right angles at \(D\); \(AD\) common \(\Rightarrow\) RHS congruent \(\Rightarrow AB=AC\), making \(\triangle ABC\) isosceles. |
| Q3. In isosceles \(\triangle ABC\), altitudes \(BE\) and \(CF\) to equal sides are drawn. Show \(BE=CF\). | Right triangles \(\triangle ABE\) and \(\triangle ACF\) with \(AB=AC\), right angles, and \(\angle A\) common \(\Rightarrow\) RHS congruence \(\Rightarrow BE=CF\). |
| Q4. In \(\triangle ABC\), altitudes \(BE\perp AC\) and \(CF\perp AB\) are equal. Show (i) \(\triangle ABE\cong \triangle ACF\); (ii) \(AB=AC\). |
(i) \(BE=CF\) (given), right angles, \(\angle A\) common \(\Rightarrow\) RHS \(\Rightarrow\) congruent. (ii) From CPCT, \(AB=AC\); hence \(\triangle ABC\) is isosceles. |
| Q5. \(\triangle ABC\) and \(\triangle DBC\) are isosceles on same base \(BC\). Show \(\angle ABD=\angle ACD\). | In isosceles \(\triangle ABC\), \(\angle ABC=\angle ACB\). Similarly, in \(\triangle DBC\), \(\angle CBD=\angle BDC\). Using linear pairs around vertex \(B\) and \(C\), compute equalities to get \(\angle ABD=\angle ACD\). |
| Q6. In isosceles \(\triangle ABC\) with \(AB=AC\), side \(BA\) is produced to \(D\) such that \(AD=AB\). Show \(\angle BCD=90^\circ\). | Construct: since \(AD=AB=AC\), \(\triangle ACD\) is isosceles with \(AC=AD\). Using exterior angle relations at \(C\) and base angle equalities, sum shows \(\angle BCD=90^\circ\). |
| Q7. Right triangle \(ABC\) with \(\angle A=90^\circ\) and \(AB=AC\). Find \(\angle B\) and \(\angle C\). | \(\angle B=\angle C\) and \(\angle B+\angle C=90^\circ\Rightarrow \angle B=\angle C=45^\circ\). |
| Q8. Show that each angle of an equilateral triangle is \(60^\circ\). | All sides equal \(\Rightarrow\) all base angles equal. Sum \(=180^\circ\Rightarrow 3\theta=180^\circ\Rightarrow \theta=60^\circ\). |
Exercise 7.3 — Textbook Questions & Perfect Solutions
| Question | Solution |
|---|---|
| Q1. \(\triangle ABC\) and \(\triangle DBC\) are isosceles on base \(BC\), with \(A,D\) on same side. If \(AD\) meets \(BC\) at \(P\), show (i) \(\triangle ABD\cong \triangle ACD\); (ii) \(\triangle ABP\cong \triangle ACP\); (iii) \(AP\) bisects \(\angle A\) and \(\angle D\); (iv) \(AP\) is perpendicular bisector of \(BC\). |
(i) Using base angles equalities in the two isosceles triangles and common side \(AD\), apply ASA to get congruence. (ii) With (i) and shared altitude/median properties, triangles about \(P\) are congruent. (iii) From CPCT, adjacent angles at \(A\) and \(D\) are bisected by \(AP\). (iv) CPCT also yields \(BP=PC\) and right angle at \(P\) due to equal adjacent angles \(\Rightarrow AP\) is perpendicular bisector of \(BC\). |
| Q2. In isosceles \(\triangle ABC\) with \(AB=AC\), altitude \(AD\). Show (i) \(AD\) bisects \(BC\); (ii) \(AD\) bisects \(\angle A\). |
(i) \(\triangle ABD\) and \(\triangle ACD\): \(AB=AC\), right angles at \(D\), \(AD\) common \(\Rightarrow\) RHS \(\Rightarrow BD=DC\). (ii) From congruence, \(\angle BAD=\angle DAC\), so \(AD\) is angle bisector. |
| Q3. Sides \(AB,BC\) and median \(AM\) of \(\triangle ABC\) are equal respectively to \(PQ,QR\) and median \(PN\) of \(\triangle PQR\). Show (i) \(\triangle ABM\cong \triangle PQN\); (ii) \(\triangle ABC\cong \triangle PQR\). |
(i) In \(\triangle ABM\) and \(\triangle PQN\): \(AB=PQ\), \(AM=PN\), and \(BM=QN\) because medians to equal sides imply halves equal: \(BM=\frac{BC}{2}=\frac{QR}{2}=QN\). So SSS \(\Rightarrow\) congruent. (ii) From CPCT, \(BC=QR\) (given), \(AC=PR\) (deduced). With \(AB=PQ\), apply SSS to whole triangles. |
| Q4. \(BE\) and \(CF\) are equal altitudes of \(\triangle ABC\). Prove \(\triangle ABC\) is isosceles. | Right triangles \(\triangle ABE\) and \(\triangle ACF\) with \(BE=CF\) and \(\angle A\) common \(\Rightarrow\) RHS congruent \(\Rightarrow AB=AC\). Hence \(\triangle ABC\) is isosceles. |
| Q5. In isosceles \(\triangle ABC\) (\(AB=AC\)), draw \(AP\perp BC\). Show \(\angle B=\angle C\). | \(\triangle ABP\) and \(\triangle ACP\): right angles at \(P\), \(AP\) common, \(AB=AC\). By RHS, triangles congruent \(\Rightarrow \angle B=\angle C\). |
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