Advertisements
Advertisements
प्रश्न
O is any point in the ΔABC such that the perpendicular drawn from O on AB and AC are equal. Prove that OA is the bisector of ∠BAC.
Advertisements
उत्तर

In ΔPOA and ΔQOA
∠OPA = ∠OQA = 90°
OP = OQ ...(given)
AO = AO
Therefore, ΔPOA ≅ ΔQOA ...(SSA criteria)
Hence, ∠PAO = ∠QAO
Thus, OA bisects ∠BAC.
APPEARS IN
संबंधित प्रश्न
If ΔABC ≅ ΔFED under the correspondence ABC ↔ FED, write all the Corresponding congruent parts of the triangles.
In a squared sheet, draw two triangles of equal areas such that
The triangles are not congruent.
What can you say about their perimeters?
In triangles ABC and PQR three equality relations between some parts are as follows:
AB = QP, ∠B = ∠P and BC = PR
State which of the congruence conditions applies:
As shown in the following figure, in ΔLMN and ΔPNM, LM = PN, LN = PM. Write the test which assures the congruence of the two triangles. Write their remaining congruent parts.

In the given figure, ∠P ≅ ∠R seg, PQ ≅ seg RQ. Prove that, ΔPQT ≅ ΔRQS.

The following figure shown a triangle ABC in which AB = AC. M is a point on AB and N is a point on AC such that BM = CN.
Prove that:
State, whether the pairs of triangles given in the following figures are congruent or not:

A is any point in the angle PQR such that the perpendiculars drawn from A on PQ and QR are equal. Prove that ∠AQP = ∠AQR.
In a triangle ABC, if D is midpoint of BC; AD is produced upto E such as DE = AD, then prove that:
a. DABD andDECD are congruent.
b. AB = EC
c. AB is parallel to EC
If AB = QR, BC = PR and CA = PQ, then ______.
