Q. 113.7( 3 Votes )

# In ΔABC, ∠B = 90° and P is the mid-point of AC. Prove that PB = PA = AC.

Answer :

In the ΔABC, P is the mid-point of AC and ∠B = 90°.

To Prove: PA = PB = AC

Construction: Draw PK ïï BC

Proof: Since KP ïï BC with transversal AB

.

Also, ∠1 + ∠2 = 180°

.

Now in ΔAPK and ΔBPK,

KP = KP...(common)

∠1 = ∠2...(prove above)

AK = KB...(since KP ïï BC and P is the mid-point of AC)

.

.

PA = AC ...( P is the mid-point of AC)

PA = PB = AC.

To Prove: PA = PB = AC

Construction: Draw PK ïï BC

Proof: Since KP ïï BC with transversal AB

.

^{.}. ∠1 = ∠B = 90°Also, ∠1 + ∠2 = 180°

.

^{.}. ∠2 = 180° - ∠1 = 180° - 90° = 90°Now in ΔAPK and ΔBPK,

KP = KP...(common)

∠1 = ∠2...(prove above)

AK = KB...(since KP ïï BC and P is the mid-point of AC)

.

^{.}. ΔAPK ≅ ΔBPK.

^{.}. PA = PB butPA = AC ...( P is the mid-point of AC)

PA = PB = AC.

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