Advertisements
Advertisements
प्रश्न
ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
Advertisements
उत्तर
We have given, ΔABC which is an isosceles right triangle with AB = AC and AD is the bisector of ∠A.
Now in ΔABC,
AB = AC ...[Given]
⇒ ∠C = ∠B ...(1) [Angles opposite to equal sides are equal]
Now, in ΔABC, ∠A = 90°
∠A + ∠B + ∠C = 180° ...[Angle sum property of Δ]
⇒ 90° + ∠B + ∠B = 180° ...[From (1)]
⇒ 2∠B = 90°
⇒ ∠B = 45°
⇒ ∠B = ∠C = 45° or ∠3 = ∠4 = 45°
Also, ∠1 = ∠2 = 45° ...[∵ AD is bisector of ∠A]
Also, ∠1 = ∠3, ∠2 = ∠4 = 45°
⇒ BD = AD, DC = AD ...(2) [Sides opposite to equal angles are equal]
Thus, BC = BD + DC = AD + AD ...[From (2)]
⇒ BC = 2AD
APPEARS IN
संबंधित प्रश्न
Which congruence criterion do you use in the following?
Given: ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ΔPQR ≅ ΔXYZ
You want to show that ΔART ≅ ΔPEN,
If you have to use SSS criterion, then you need to show
1) AR =
2) RT =
3) AT =
You want to show that ΔART ≅ ΔPEN,
If it is given that AT = PN and you are to use ASA criterion, you need to have
1) ?
2) ?
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.
Use the information in the given figure to prove:
- AB = FE
- BD = CF
The following figure shows a circle with center O.
If OP is perpendicular to AB, prove that AP = BP.
A triangle ABC has ∠B = ∠C.
Prove that: The perpendiculars from B and C to the opposite sides are equal.
A line segment AB is bisected at point P and through point P another line segment PQ, which is perpendicular to AB, is drawn. Show that: QA = QB.
In quadrilateral ABCD, AD = BC and BD = CA.
Prove that:
(i) ∠ADB = ∠BCA
(ii) ∠DAB = ∠CBA
In the following figure, OA = OC and AB = BC.
Prove that: ΔAOD≅ ΔCOD
