Advertisements
Advertisements
प्रश्न
In the parallelogram ABCD, the angles A and C are obtuse. Points X and Y are taken on the diagonal BD such that the angles XAD and YCB are right angles.
Prove that: XA = YC.
Advertisements
उत्तर
ABCD is a parallelogram in which ∠A and ∠C are obtuse.
Points X and Y are taken on the diagonal BD.
Such that ∠XAD = ∠YCB = 90°.
We need to prove that XA = YC
Proof:
ln ΔXAD and ΔYCB
∠XAD = ∠YCB= 90° ...[ Given ]
AD = BC ...[ Opposite sides of a parallelogram ]
∠ADX = ∠CBY ...[ Alternate angles ]
∴ By Angle-Side-Angle criterion of congruence,
ΔXAD ≅ ΔYCB
The corresponding parts of the congruent triangles are congruent.
∴ XA = YC ...[ c.p.c.t. ]
Hence proved.
APPEARS IN
संबंधित प्रश्न
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.
In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
In the given figure, prove that:
CD + DA + AB + BC > 2AC
In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD. Prove that:
AB = CE.
A triangle ABC has ∠B = ∠C.
Prove that: The perpendiculars from B and C to the opposite sides are equal.
The perpendicular bisectors of the sides of a triangle ABC meet at I.
Prove that: IA = IB = IC.
In the following figure, AB = AC and AD is perpendicular to BC. BE bisects angle B and EF is perpendicular to AB.
Prove that: BD = CD
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.
PQRS is a parallelogram. L and M are points on PQ and SR respectively such that PL = MR.
Show that LM and QS bisect each other.
Which of the following is not a criterion for congruence of triangles?
