practice problems pending

Q1. The side AB of a parallelogram ABCD is produced both ways to F and G, so that AF = AD and BG = BC. Prove that FD and GC produced intersect at right angle.
S1.
There are 2 ways to draw the diagram here.
Both give us the desired proof.


Q2.
In triangle AQB, points P and D lie on sides AB and AQ, respectively, such that

[APQ]=[ABD].

Through D, draw the line DR∥AB, meeting BQ at R.
Through B draw a line || to AQ which meets DR at C.
Pr. th.
RC=AP.
S2.


ABCD is a ||gram.
AB = CD____[1]
AD = BC

Given [APQ] = [ABD]
=> Sin(A).AP.AQ = Sin(A).AB.AD
=> AP/AB = AD/AQ
=> Triangles APD and ABQ are similar by SAS.
=> PD || BR
=> BPDR is a ||gram
=> PD = BR and BP = DR___[2]
From [1]:
AP + PB = DR + RC
Using [2]:
AP = RC H.P.

Q3.
ABCD is a parallelogram. Through C, a straight line RQ is drawn outside the parallelogram, and AP, BQ, DR are drawn perpendicular to RQ. Show that

DR + BQ = AP.


S3. 



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