# Congurency In Triangles

## Exercise 7.3

Question 1: Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC . If AD is extended to intersect BC at P, show that

1. ΔABD≅ ΔACD
2. ΔABP≅ ΔACP
3. AP bisects ∠A as well as ∠D
4. AP is perpendicular bisector of BC

AB=AC
BD=CD
AD=AD
So, ΔABD≅ ΔACD (SSS rule)

In ΔABP and ΔACP
AB=AC
AP=AP
∠ABP=∠ACP (angles opposite to equal sides)
So, ΔABP≅ ΔCP (SAS rule)
Since ΔABP≅ ΔACP
So, ∠BAP=∠CAP
So, AP is bisecting ∠BAC
Similarly, ΔBDP and ΔCDP can be proven to be congruent and as a result it can be proved that AP is bisecting ∠BDC

Question 2: AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

Answer: This can be solved like previous question.

Question 3: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR. Show that:

• ΔABM≅ ΔPQN
• ΔABC≅ ΔPQR

AB=PQ
AM=PN
BM=QN (median bisects the base)
So, ΔABM≅ ΔPQN

In ΔABC and ΔPQR
AB=PQ
BC=QR
AC=PR (equal medians mean third side will be equal)
So, ΔABC≅ ΔPQR

Question 4: BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

BE=CE (perpendicular)
AB=BC (hypotenuse)
So, ΔAEB≅ ΔAFC

Question 5: ABC is an isosceles triangle with AB = AC. Draw AD ┴ BC to show that angle B = angle C.

AC=AB
AD=AD
∠ADC=∠ADB
So, ΔADC≅ ΔADB
So, ∠ACD=∠ABC