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In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that: (i) Δ APB is similar to Δ CPD. (ii) PA x PD = PB x PC.
In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that: (i) Δ APB is similar to Δ CPD. (ii) PA x PD = PB x PC.
Class 10 | Exercise 15A | ICSE | Maths | Selina | Similarity (With Applications to Maps and Models)
In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that: (i) Δ APB is similar to Δ CPD. (ii) PA x PD = PB x PC.

Selina Solutions Concise Class 10 Maths Chapter 15 ex. 15(A) - 2

(I) In\[\vartriangle APB\text{ }and\text{ }\vartriangle CPD\], we have

\[\angle APB\text{ }=\angle CPD\] [Vertically inverse angles]

\[\angle ABP\text{ }=\angle CDP\] [Alternate points as, AB||DC]

Consequently, \[\vartriangle APB\text{ }\sim\text{ }\vartriangle CPD\]by \[AA\] similarity criterion.

 

(ii) As \[\vartriangle APB\text{ }\sim\text{ }\vartriangle CPD\]

Since the comparing sides of comparative triangles are relative, we have

\[PA/PC\text{ }=\text{ }PB/PD\]

Consequently,

\[PA\text{ }x\text{ }PD\text{ }=\text{ }PB\text{ }x\text{ }PC\]

i

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