Question

plz help asap
$( \sqrt{} (thanks \: in \: advance)) {}^{2}$

Answer 1

Given ABCD is a parallelogram,PQ perpendicular to CD , CE perpendicular to AB.

RTP:ar(CPB)+ar(APD)=ar(PCD)

Contruction:Draw MD perpendicular to AP

Proof:In triangle PQC and PEC

1)angleEPC=angleQCP(since AB parallel to CD:alternate interior angles by taking PC as transversal)

2)PC=PC(common side)

3)angleQPC=angleECP(since AB parallel to CD:alternate interior angles by taking PC as transversal)

Therefore by ASA congruency triangle PQC is congruent to triangle PEC

By CPCT

PQ=EC

QC=EP

Similarly

triangleMPD is congruent to triangle PDQ

By CPCT

MD=PQ

MP=DQ

=>ar(PBC)=

$\frac{1}{2} \times ep \times pb$

Since EP=QC

=> PB=QC(PB is a part of EP)

and PQ=EC

ar(PBC)=

$\frac{1}{2} \times qc \times pq$

But

$\frac{1}{2} \times qc \times pq$

is equal to ar(PQC)

=>ar(PBC)=ar(PQC) -> 1

similarly

ar(APD)=ar(PQC) -> 2

if we add 1 and 2

ar(PBC) + ar(APD) = 2 * ar(PQC)

=>ar(CPB) + ar(APD) = ar(PCD)

HENCE PROVED