Question

In fig seg AC & seg BD intersect each other at p
$\frac{ap}{pc } = \frac{bp}{pd} then \: prove \: th at \: abp \: dpc$
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Answer 1

Answer:

according to question

Step-by-step explanation:

ap/pc=bp/pd

and angleAPB=CPD

SO BY SAS rule triangle APB IS SIMILAR TO TRIANGLE CPD