ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (See the given figure). Show that (i) ?APB ? ?CQD (ii) AP = CQ

Asked by adityaprakhar | 25th Oct, 2010, 08:13: PM

Expert Answer:

Dear Student,
 
In triangle CQB,
angle QCB + angle QBC = 90                (1)
 
and in rectangle ABCD, angle BCD = 90
=> angle BCQ + angle QCD = 90            (2)
 
From (1) and (2) we get,
 
angle QBC = angle QCD
 
 
Similarly we can prove that angle PDA = angle PAB (You are encouraged to develop this part on your own based on the above reasoning.)
 
Also, since angle QBC and angle PDA are interior opposite angles, therefore
 
angle QBC = angle PDA
 
=> angle QCD = angle PAB    (3)
 
 
Now, consider triangle PAB and trianlge QCD,
 
angle PAB = angle QCD    (from (3))
angle APB = angle CQD    (both 90)
angle ABP = angle CDQ    (remaining angle in the triangle)
side AB = side CD            (opposite sides of rectangle)
 
Therefore by ASA congruence criterion, the two triangles are congruent.
 
 
For second part,
since the two triangles are congruent, corresponding sides are also equal.
Therefore, side AP = side CQ.
 
 
Regards Topperlearning.

Answered by  | 26th Oct, 2010, 01:05: AM

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