Question
ABC and ADC are two right triangles with common hypotenuse AC. Prove that $\angle C A D = \angle C B D.$
✓
Answer
Given: ABC and ADC are two right triangles with common hypotenuse AC.
To prove: $\angle C A D = \angle C B D$
Proof : AC is the common hypotenuse and ABC and ADC are two right triangles.
$\therefore \angle A B C = 90 ^ { \circ } = \angle A D C$
$\Rightarrow$ Both the triangles are in the same semi-circle
$\therefore$ Points A, B, D and C are concyclic
$\therefore$DC is a chord
$\therefore \angle C A D = \angle C B D$ |$\because$ Angles in the same segment are equal.
To prove: $\angle C A D = \angle C B D$
Proof : AC is the common hypotenuse and ABC and ADC are two right triangles.
$\therefore \angle A B C = 90 ^ { \circ } = \angle A D C$
$\Rightarrow$ Both the triangles are in the same semi-circle
$\therefore$ Points A, B, D and C are concyclic
$\therefore$DC is a chord
$\therefore \angle C A D = \angle C B D$ |$\because$ Angles in the same segment are equal.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the values of x in each of the following:$5^{\text{x}-2}\times3^{2\text{x}-3}=135$
→At a Ramzan Mela, a stall keeper in one of the food stalls has a large cylindrical vessel of the base radius $15 \ cm$ filled upto a height of $32 \ cm$ with the orange juice. The juice is filled in small cylindrical glasses $($see figure$)$ of radius $3\ cm$ upto a height of $8 \ cm$ and sold for $Rs.15$ each. How much money does the stall keeper receive by selling the juice completely?
→In figure ABC and BDE are two equilateral triangles such that D is the midpoint of BC. AE intersects BC in F. Prove that:
$\text{ar}(\triangle\text{ABC})=2\text{ar}(\triangle\text{BEC})$
→$\text{ar}(\triangle\text{ABC})=2\text{ar}(\triangle\text{BEC})$
If x = 3 and y = -1, find the values of the following using in identity:
$\Big(\frac{5}{\text{x}}+5\text{x}\Big)\Big(\frac{25}{\text{x}^2}-25+25\text{x}^2\Big)$
→$\Big(\frac{5}{\text{x}}+5\text{x}\Big)\Big(\frac{25}{\text{x}^2}-25+25\text{x}^2\Big)$
How many cubic centimeters of iron are there in an open box whose external dimensions are $36\ cm, 25\ cm$ and $16.5\ cm,$ the iron being $1.5\ cm$ thick throughout? If $1$ cubic $\ cm$ of iron weighs $15\ gms.$ Find the weight of the empty box in $\ kg.$
→Factorize : $x^4 + x^2 + 25.$
→Factorise $: 9(2a - b)^2 - 4(2a - b) - 13$
→The following are the marks (out of 100) of 60 students in mathematics:
16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35, 72, 55, 75, 31, 52, 28,72, 97, 74, 45, 62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92, 62, 52, 56, 15, 63,25, 36, 54, 44, 47, 27, 72, 17, 4, 30
Construct a grouped frequency distribution table with width 10 of each class starting from 0-9.
16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35, 72, 55, 75, 31, 52, 28,72, 97, 74, 45, 62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92, 62, 52, 56, 15, 63,25, 36, 54, 44, 47, 27, 72, 17, 4, 30
Construct a grouped frequency distribution table with width 10 of each class starting from 0-9.
Simplify:
$\frac{(25)^{\frac{3}{2}}\times(243)^{\frac{3}{5}}}{(16)^{\frac{5}{4}}\times(8)^{\frac{4}{3}}}$
→$\frac{(25)^{\frac{3}{2}}\times(243)^{\frac{3}{5}}}{(16)^{\frac{5}{4}}\times(8)^{\frac{4}{3}}}$
If $a + b + c = 0$ and $a^2 + b^2 + c^2 = 16,$ find the value of $ab + bc + ca.$
→