Question

In adjoining figure, AB is a line segment and P its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB. Show that
(i) ∆DAP ≅ ∆EBP
(ii) AD = BE

Answer 1

$\huge\star{\green{\underline{\underline{\mathfrak{Explanation:}}}}}$

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In the question, it is given that P is the mid-point of line segment AB.

Also, ∠BAD = ∠ABE and ∠EPA = ∠DPB

(i) It is given that ∠EPA = ∠DPB

Now, add ∠DPE om both sides.

∠EPA + ∠DPE = ∠DPB + ∠DPE

This implies that angles DPA and EPB are equal i.e. ∠DPA = ∠EPB

Now, consider the triangles DAP and EBP.

∠DPA = ∠EPB

AP = BP (Since P is the mid-point of the line segement AB)

∠BAD = ∠ABE (As given in the question)

So, by ASA congruency, ΔDAP ≅ ΔEBP.

(ii) By the rule of CPCT, AD = BE.

$\large\star{\purple{\underline{\mathfrak{Hence,peoved!}}}}$

Answer 2

(i) ∆DAP ≅ ∆EBP

(ii) AD = BE

Explanation:

Given: AB is a line segment and P its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB.  

To prove:  

(i) ∆DAP ≅ ∆EBP

(ii) AD = BE

Proof:

In ∆DAP and ∆EBP, we know that AP = PB as P is the midpoint of AB.

Given that  ∠BAD = ∠ABE.

∠EPA + ∠EPD + ∠DPB = 180

Given ∠EPA = ∠DPB, so ∠EPD + 2∠DPB = 180

∠EPD = 180 - 2∠DPB

In ∆DAP,

∠DPA = ∠EPD + ∠EPA

In ∆EBP

∠EPB = ∠EPD + ∠DPB

Given that ∠EPA = ∠DPB, so we get ∠DPA = ∠EPB.

Two pairs of angles are equal.

So ∆DAP ≅ ∆EBP  by ASA. Hence proved.

Since the triangles are congruent, the corresponding sides are equal. So  AD = BE. Hence proved.