Advertisements
Advertisements
Question
In Fig. 10.99, AD ⊥ CD and CB ⊥. CD. If AQ = BP and DP = CQ, prove that ∠DAQ = ∠CBP.
Advertisements
Solution
Given that, in the figure AD ⊥ CD and CB ⊥ CD and AQ = BP,DP =CQ
We have to prove that ∠DAQ=∠CBP
Given that DP= QC
Add PQ on both sides
Given that DP=QC
Add PQ on both sides
⇒ DP+PQ=PQ+QC
⇒ DQ=PC ................(1)
Now, consider triangle DAQ and CBP,
We have
∠ADQ=∠BCP=90° [given]
AQ=BP [given]
And DQ=PC [given]
So, by RHS congruence criterion, we have ΔDAQ≅ΔCBP
Now,
∠DAQ=∠CBP [ ∵Corresponding parts of congruent triangles are equal]
∴ Hence proved
APPEARS IN
RELATED QUESTIONS
AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.

Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:
- ΔAPB ≅ ΔAQB
- BP = BQ or B is equidistant from the arms of ∠A.

Which congruence criterion do you use in the following?
Given: ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ΔPQR ≅ ΔXYZ

If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

Which of the following statements are true (T) and which are false (F):
Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal equal to the hypotenuse and a side of the other triangle.
If the following pair of the triangle is congruent? state the condition of congruency :
In Δ ABC and Δ DEF, AB = DE, BC = EF and ∠ B = ∠ E.
In the given figure, AB = DB and Ac = DC.

If ∠ ABD = 58o,
∠ DBC = (2x - 4)o,
∠ ACB = y + 15o and
∠ DCB = 63o ; find the values of x and y.
In ∆ABC, AB = AC. Show that the altitude AD is median also.
A point O is taken inside a rhombus ABCD such that its distance from the vertices B and D are equal. Show that AOC is a straight line.
In ΔABC, AB = AC and the bisectors of angles B and C intersect at point O.
Prove that : (i) BO = CO
(ii) AO bisects angle BAC.
