Question: In the rhombus ABCD, AP is constructed perpendicular to BC and intersects the diagonal BD at Q. How to work this out? (working outs + reasons and explanations)

(a) State why 'angle ADB' = 'angle CDB'enter image source here .

(b) Prove that △AQD ≡ △CQD.

(c) Show that 'angle DAQ' is a right angle.

(d) Hence find 'angle QCD'.

1 Answer
Mar 17, 2018

See explanation.

Explanation:

enter image source here
Some of the properties of a rhombus :
1) all sides are congruent, #=> AB=BC=CD=DA#,
2) opposite angles are congruent, #=> angleADC=angleABC#,
3) opposite sides are parallel, #=> AD# // #BC, and AB# // #DC#,
4) the diagonals bisect the angles, #=> angleADB=angleCDB=1/2angleADC#

a) #BD# bisects #ADC, => angleADB=angleCDB#,

b) #AD=CD, DQ# is common side, and #angleADB=angleCDB#,
#=> DeltaAQD and DeltaCQD# are congruent.

c) as #AD# // #BC#, and given that #angleBPA=90^@#,
#=> angleDAQ=angleBPA=90^@#

d) as #DeltaQCD and DeltaQAD# are congruent,
#=> angleQCD=angleQAD=90^@#