Question

AB is a line segment and P is the mid point. D and
E are points on the same side of AB such that
<BAD = <ABE and < EPA = <DPB
(see Fig. 7.22). Show that
i)∆DAP=~∆ EBP
ii) AD=BE
​

Answer 1

Step-by-step explanation:

I HOPE THAT THIS ANSWER IS HELPFUL TO ALL

Answer 2

Answer:

Given

• AB is a line segment and P is the midpoint.
• <BAD = <ABE and <EPA = <DPB

To show

• i) triangle (DAP) =~ triangle(EBP)
• ii) AD = BE

Solution

P is the midpoint of AB.

So, AP = PB equation i)

<APD = <EPA + <DPE equation ii)

and <BPE = <DPB + <DPE equation iii)

Though, <EPA = <DPE given

Hence, <APD = <BPE equation iv)

Now, in triangle (DAP) and triangle(EBP)

<BAD or <PAD = <PBE or <ABE (Angle) given

AP = PB (Side) by equation i)

<APD = <BPE (angle) by equation iv)

Now, by ASA criterion rule

i) triangle(DAP)=~ triangle (EBP)

ii) by CPCT AD = BE