Question

In the figure triangle ADB and triangle CDB are drawn on the same base BD. If AC and BD intersect at O,then prove that A(triangle ADB) /A(triangle CDB) =AO/CO.

Answer 1

Proof:

Given that ΔADB and ΔCDB are drawn on the same base BD.

The lines AC and BD intersect at O.

To prove: $\frac{A(\triangle A D B)}{A(\triangle C D B)}=\frac{A O}{C O}$

Let us draw AN ⊥ BD and CM ⊥ BD

The area of ΔADB is given by

$A(\triangle {ADB})=\frac{1}{2} \times B D\times A N$

The area of ΔCDB is given by

$A( \triangle{CBD})=\frac{1}{2} \times D B \times C M$

Taking the ratio, we get,

$\frac{A(\triangle A DB )}{A( \triangle CBD)}=\frac{\frac{1}{2} \times BD \times A N}{\frac{1}{2} \times DB \times CM}$

Cancelling the terms, we get,

$\frac{A(\triangle A DB )}{A( \triangle CBD)}=\frac{ A N}{ CM}$  -----------(1)

Now, let us consider the ΔAON and ΔCOM

$\angle {ANO}=\angle {CMO}$  (right angles)

$\angle {AON}=\angle {COM}$  (vertically opposite angles)

Then, by AA similarity criterion, we get,

$\triangle AON \sim \triangle COM$

Also, if 2 triangles are similar then their corresponding sides are in the same ratio.

Thus, we have,

$\frac{AN}{CM} =\frac{AO}{CO}$ -------(2)

From the equations (1) and (2), we get,

$\frac{A(\triangle A D B)}{A(\triangle C D B)}=\frac{A O}{C O}$

Hence proved

Learn more:

(1) In the figure triangle ADB and triangle cdb are on the same base DB if AC and BD intersect at O then prove that area of triangle ADB upon area of triangle CBD is equals AO upon CO​

brainly.in/question/15521174

(2) In the figure triangle adb and triangle cdb are drawn on the same base bd. if ac and bd intersects at o,then prove that a (triangle adb)/a(triangle cdb) = ao/co​

brainly.in/question/15424891

Answer 2

Proof:

Given that ΔADB and ΔCDB are drawn on the same base BD.

The lines AC and BD intersect at O.

To prove: $\frac{A(\triangle A D B)}{A(\triangle C D B)}=\frac{A O}{C O}$

Let us draw AN ⊥ BD and CM ⊥ BD

The area of ΔADB is given by

$A(\triangle {ADB})=\frac{1}{2} \times B D\times A N$

The area of ΔCDB is given by

$A( \triangle{CBD})=\frac{1}{2} \times D B \times C M$

Taking the ratio, we get,

$\frac{A(\triangle A DB )}{A( \triangle CBD)}=\frac{\frac{1}{2} \times BD \times A N}{\frac{1}{2} \times DB \times CM}$

Cancelling the terms, we get,

$\frac{A(\triangle A DB )}{A( \triangle CBD)}=\frac{ A N}{ CM}$  -----------(1)

Now, let us consider the ΔAON and ΔCOM

$\angle {ANO}=\angle {CMO}$  (right angles)

$\angle {AON}=\angle {COM}$  (vertically opposite angles)

Then, by AA similarity criterion, we get,

$\triangle AON \sim \triangle COM$

Also, if 2 triangles are similar then their corresponding sides are in the same ratio.

Thus, we have,

$\frac{AN}{CM} =\frac{AO}{CO}$ -------(2)

From the equations (1) and (2), we get,

$\frac{A(\triangle A D B)}{A(\triangle C D B)}=\frac{A O}{C O}$

Hence proved

Learn more:

(1) In the figure triangle ADB and triangle cdb are on the same base DB if AC and BD intersect at O then prove that area of triangle ADB upon area of triangle CBD is equals AO upon CO

brainly.in/question/15521174

(2) In the figure triangle adb and triangle cdb are drawn on the same base bd. if ac and bd intersects at o,then prove that a (triangle adb)/a(triangle cdb) = ao/co

brainly.in/question/15424891