Question

In the adjoining figure, AABC is right-angled at B and P is the
mid-point of AC. Show that, PA = PB = PC.
​

Answer 1

Answer:

hiii friend here is ur solution

Given, in triangle ABC p is the midpoint of AC

therefore , AP=AC

and QP llel to BC

now , in tri APB and tri PCB

= AP=PC ( since p is the mid point of AC)

= angle APB=angle BPC ( alt. in angle since QR llel to BC)

= BP=BP ( common)

since tri APB is congruence to tri PCB

Hence PB =AP ( c.p.c.t)

therefore , AP =BP=PC .

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Answer 2

Answer:

Given, in triangle ABC p is the midpoint of AC

therefore , AP=AC

and QP llel to BC

now , in tri APB and tri PCB

= AP=PC ( since p is the mid point of AC)

= angle APB=angle BPC ( alt. in angle since QR llel to BC)

= BP=BP ( common)

since tri APB is congruence to tri PCB

Hence PB =AP ( c.p.c.t)

therefore , AP =BP=PC .

Step-by-step explanation:

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