Question

"Question 7 AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that (i) ΔDAP ≅ ΔEBP (ii) AD = BE

Class 9 - Math - Triangles Page 120"

Answer 1

ASA(angle side angle):

Two Triangles are congruent if two angles and the included side of One triangle are equal to two angles & the included side of the  other triangle.

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Given:

P is mid-point of AB i.e, AP=BP

∠BAD = ∠ABE and ∠EPA = ∠DPB

 

To Prove:

i)   ΔDAP ≅ ΔEBP

ii)   AD=BE

 

Proof:

We have

(i)  ∠EPA = ∠DPB (Adding ∠DPE both sides)

∠EPA + ∠DPE = ∠DPB + ∠DPE

⇒ ∠DPA = ∠EPB

In ΔDAP & ΔEBP,
∠DPA = ∠EPB                                          (proved above)

AP = BP                                            (P is mid-point of AB)
∠BAD = ∠ABE                                   (Given)

Hence, ΔDAP ≅ ΔEBP           (by ASA congruence rule.)

(ii) Since , ΔDAP ≅ ΔEBP

Then,AD = BE

( by CPCT)

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Answer 2

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