Question

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' .

(b) Prove that △AQD ≡ △CQD.

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

(d) Hence find 'angle QCD'.

Answer 1

See explanation.

Explanation:

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^@`