Advertisements
Advertisements
प्रश्न
On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that ∠BAC = ∠BDC.
Advertisements
उत्तर
Given: ΔACB and ΔADB are two right angled triangles with common hypotenuse AB.
To prove: ∠BAC = ∠BDC
Construction: Join CD.
Proof: Let O be the mid-point of AB
Then, OA = OB = OC = OD.
Since, mid-point of the hypotenuse of a right triangle is equidistant from its verticles.
Now, draw a circle to pass through the points A, B, C and D with O as centre and radius equal to OA.
We know that, angles in the same segment of a circle are equal.
From the figure, ∠BAC and ∠BDC are angles of same segment BC.
∴ ∠BAC = ∠BDC
Hence proved.
APPEARS IN
संबंधित प्रश्न
In Fig. 1, PA and PB are tangents to the circle with centre O such that ∠APB = 50°. Write the measure of ∠OAB.
In the given figure, PQ is a chord of length 8cm of a circle of radius 5cm. The tangents at P and Q intersect at a point T. Find the length TP
In the given figure, PO⊥QO. The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are right bisector of each other.
In following fig. ABC is an equilateral triangle . A circle is drawn with centre A so that ot cuts AB and AC at M and N respectively. Prove that BN = CM.
ABC is a right triangle in which ∠B = 90°. If AB = 8 cm and BC = 6 cm, find the diameter of the circle inscribed in the triangle.
The radius of a circle of diameter 24 cm is _______
In the adjoining figure, seg DE is the chord of the circle with center C. seg CF⊥ seg DE and DE = 16 cm, then find the length of DF?
The length of the tangent from point A to a circle, of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is ______
From the figure, identify three radii.
What is the name for the perimeter of a circle?
