Home
Class 9
MATHS
ABC and ADC are two right triangles wit...

ABC and ADC are two right triangles with common hypotenuse AC. Prove that `/_C A D\ =/_C B Ddot`

Text Solution

Verified by Experts

Since, `/_ABC` and `/_ADC` are 90.
These angles are in the semi-circle.
Thus, both the triangles are lying in the semi-circle and `AC` is the diameter of the circle.
Points `A`,`B`,`C` and `D` are concyclic.
Thus, `CD` is the chord.
Angles in the same segment of the circle are equal. Thus,
`/_CAD=/_CBD`
Promotional Banner

Topper's Solved these Questions

  • CIRCLES

    NCERT|Exercise EXERCISE 10.2|2 Videos
  • CIRCLES

    NCERT|Exercise Exercise 10.4|6 Videos
  • AREAS OF PARALLELOGRAMS AND TRIANGLES

    NCERT|Exercise EXERCISE 9.1|1 Videos
  • CONSTRUCTIONS

    NCERT|Exercise EXERCISE 11.2|5 Videos

Similar Questions

Explore conceptually related problems

ABC and ADC are two right triangles with common hypotenuse AC. Prove that /_CAD=/_CBD

ABC AND ADC are two right triangles with common hypotenuse AC. Prove that /_CAD=/_CBD

In the given figure, triangles ABC and DCB are right angled at A and D respectively and AC = DB. Prove that /_\ ABC = /_\ DCB .

In figure ABC and AMP are two right triangles, right angles at B and M respectively. Prove that(i) DeltaA B C~ DeltaA M P (ii) (C A)/(P A)=(B C)/(M P)

ABC and ADC are two equilateral triangles on a common base AC.Find the angles of the resulting quadrilateral.Show that it is a rhombus.

ABC is an isosceles right triangle right- angled at C. Prove that AB^(2)=2AC^(2)

ABC is a right triangle right-angled at Cand AC=sqrt(3)BC. Prove that /_ABC=60^(@).

In right angled triangle ABC,/_A is right angle.AD is perpendicular to the hypotenuse BC.Prove that (Delta ABC)/(Delta ACD)=(BC^(2))/(AC^(2))

Let ABC be a right triangle with length of side AB=3 and hypotenuse AC=5. If D is a point on BC such that (BD)/(DC)=(AB)/(AC), then AD is equal to

In the given figure, triangles ABC and DCB are right angled at A and D respectively and AC = DB, then prove that angleACB = angleDBC .