Advertisements
Advertisements
प्रश्न
In quadrilateral ACBD, AC = AD and AB bisects ∠A (See the given figure). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?
Advertisements
उत्तर
Given: In quadrilateral ABCD, AC = AD and AB bisects ∠A.
To prove: △ABC ≌ △ABD
Proof: In △ABC and △ABD,
AC = AD ...[Given]
∠BAC = ∠BAD ...[∵ AB bisects ∠A]
AB = AB ...[Common]
∴ △ABC ≌ △ABD ...[By SAS congruence rule]
Hence, BC = BD ...[Corresponding parts of congruent triangles]
APPEARS IN
संबंधित प्रश्न
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (See the given figure). Prove that
- ΔABD ≅ ΔBAC
- BD = AC
- ∠ABD = ∠BAC.
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.
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that
- ΔDAP ≅ ΔEBP
- AD = BE
Which congruence criterion do you use in the following?
Given: EB = DB
AE = BC
∠A = ∠C = 90°
So, ΔABE ≅ ΔCDB
In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
If the following pair of the triangle is congruent? state the condition of congruency:
In ΔABC and ΔQRP, AB = QR, ∠B = ∠R and ∠C = P.
In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD. Prove that:
AB is parallel to EC.
The perpendicular bisectors of the sides of a triangle ABC meet at I.
Prove that: IA = IB = IC.
In quadrilateral ABCD, AD = BC and BD = CA.
Prove that:
(i) ∠ADB = ∠BCA
(ii) ∠DAB = ∠CBA
