Question

ABCD is a square and p is any point on of its diagonals BD. Is ΔABP ~ ΔCBP. Give reason

Answer 1

Answer:

Step-by-step explanation:

Sol: Given: ABCD be a parallelogram whose diagonal intersect at O.

To Prove: (1) ar(ΔADO) = ar(ΔCDO) (2) ar(ΔABP) = ar(ΔCBP)

Proof : AC is the diagonal of the parallelogram. ∴ AO = OC [Diagonal of a parallelogram bisect each other] O is the midpoint of AC Þ DO is the median.

⇒ ar(ΔADO) = ar(ΔCDO) [Median of a triangle divides it into two triangles of equal area.] --------- (1)

Similarly BO is the median of ΔABC. ⇒ ar(ΔABO) = ar(ΔBCO) [Median of a triangle divides it into two triangles of equal area.] --------- (2)

P is a point on BO. Therefore, OP is also the median of ΔAPC. ⇒ ar(ΔOPA) = ar(ΔOPC) --------- (3)

Subtracting equation (3) from equation (3), we get,

ar(ΔABO) − ar(ΔOPA) = ar(ΔBCO) − ar(ΔOPC) ⇒ ar(ΔABP) = ar(ΔCBP)

Hence proved

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