Answer:
In △APB and △DPC, we have
∠A=∠D=90
∘
and
∠APB=∠DPC [Vertically opposite angles]
Thus, by AA-criterion of similarity, we have
△APB∼△DPC
⇒
DP
AP
=
PC
PB
⇒ AP×PC=DP×PB [Hence proved]
Solve any question of Triangles with:-
_______
Hey I can report ummm
u-know-who's answers right?
btw running out of qns
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Answers & Comments
Answer:
In △APB and △DPC, we have
∠A=∠D=90
∘
and
∠APB=∠DPC [Vertically opposite angles]
Thus, by AA-criterion of similarity, we have
△APB∼△DPC
⇒
DP
AP
=
PC
PB
⇒ AP×PC=DP×PB [Hence proved]
Solve any question of Triangles with:-
Verified answer
In △APB and △DPC, we have
∠A=∠D=90
∘
and
∠APB=∠DPC [Vertically opposite angles]
Thus, by AA-criterion of similarity, we have
△APB∼△DPC
⇒
DP
AP
=
PC
PB
⇒ AP×PC=DP×PB [Hence proved]
_______
Hey I can report ummm
u-know-who's answers right?
btw running out of qns