Question

Que: 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. Show that

(i) ΔDAP ≅ ΔEBP

(ii) AD = BE​

Answer 1

Solution:

Given, P is the mid-point of line segment AB.

Also, ∠BAD = ∠ABE and ∠EPA = ∠DPB

(i) Given, ∠EPA = ∠DPB

Now, add ∠DPE on 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 segment AB)

∠BAD = ∠ABE (given)

So, by ASA congruency, ΔDAP ≅ ΔEBP.

(ii) By the rule of CPCT,

AD = BE

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Hope it will be helpful :)

Answer 2

$\huge\red{\underline{{\boxed{\textbf{QUESTION}}}}}$

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. Show that

(i) ΔDAP ≅ ΔEBP

(ii) AD = BE

$\huge\red{\underline{{\boxed{\textbf{ANSWER}}}}}$

It is given that ∠EPA = ∠DPB

⇒ ∠EPA + ∠DPE = ∠DPB + ∠DPE

⇒ ∠DPA = ∠EPB

InDAP andEBP,

∠DAP = ∠EBP (Given)

AP = BP (P is mid-point of AB)

∠DPA = ∠EPB (From above)

∴ ΔDAP ≅ ΔEBP (ASA congruence rule)

∴ AD = BE (By CPCT)

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@HarshPratapSingh