Since both line segments EF and BD are perpendicular to the same line segment CD, they must be parallel to each other (EF BD). Since points B, D, and A lie on the same straight line, EF is also parallel to AD.

Since both line segments EF and BD are perpendicular to the same line segment CD, they must be parallel to each other (EF BD). Since points B, D, and A lie on the same straight line, EF is also parallel to AD.

In CFE and CDB, C is common to both triangles, and CFE = CDB = 90^°. Therefore, by the Angle-Angle (AA) similarity criterion, CFE CDB. From the proportionality of corresponding sides, we have (CF)/(CD) = (EF)/(BD).

In CFE and CDB, C is common to both triangles, and CFE = CDB = 90^°. Therefore, by the Angle-Angle (AA) similarity criterion, CFE CDB. From the proportionality of corresponding sides, we have (CF)/(CD) = (EF)/(BD).

In EFG and ADG, EGF = AGD because they are vertically opposite angles. Furthermore, since EF AD, the alternate interior angles FEG and DAG are equal (as are the right angles EFG = ADG = 90^°). By the AA similarity criterion, EFG ADG, which gives the ratio (FG)/(DG) = (EF)/(AD).

In EFG and ADG, EGF = AGD because they are vertically opposite angles. Furthermore, since EF AD, the alternate interior angles FEG and DAG are equal (as are the right angles EFG = ADG = 90^°). By the AA similarity criterion, EFG ADG, which gives the ratio (FG)/(DG) = (EF)/(AD).

In the given figure, $BD \perp CD$ and $EF \perp CD$ . $AE$ intersects $CD$ at $G$ , with point $E$ lying on $BC$ and $A$ being on the extension of $BD$ (or $B-D-A$ ). Prove that $\frac{CF}{CD} = \frac{FG}{DG}$ . (CBSE-2023) Figure details: - Right-angled triangle $\triangle BDC$ with the right angle at $D$ ( $\angle BDC = 90^\circ$ ). - Line segment $EF$ is perpendicular to $CD$ at point $F$ ( $\angle EFC = 90^\circ$ ), where $E$ lies on hypotenuse $BC$ and $F$ lies on side $CD$ . - Points $B, D, A$ lie on a line such that $BD \perp CD$ and $DA \perp CD$ . - A line segment connects $E$ to $A$ , intersecting the line segment $CD$ at point $G$ , so $C, F, G, D$ are collinear in that order.

Answer: Hence proved that $\frac{CF}{CD} = \frac{FG}{DG}$ .

Step-by-step solution

Step 1: Establish parallel lines from perpendiculars to the same line

Since both line segments $EF$ and $BD$ are perpendicular to the same line segment $CD$ , they must be parallel to each other ( $EF \parallel BD$ ). Since points $B$ , $D$ , and $A$ lie on the same straight line, $EF$ is also parallel to $AD$ .

Step 2: Prove similarity of triangle CFE and triangle CDB

In $\triangle CFE$ and $\triangle CDB$ , $\angle C$ is common to both triangles, and $\angle CFE = \angle CDB = 90^\circ$ . Therefore, by the Angle-Angle (AA) similarity criterion, $\triangle CFE \sim \triangle CDB$ . From the proportionality of corresponding sides, we have $\frac{CF}{CD} = \frac{EF}{BD}$ .

Step 3: Prove similarity of triangle EFG and triangle ADG

In $\triangle EFG$ and $\triangle ADG$ , $\angle EGF = \angle AGD$ because they are vertically opposite angles. Furthermore, since $EF \parallel AD$ , the alternate interior angles $\angle FEG$ and $\angle DAG$ are equal (as are the right angles $\angle EFG = \angle ADG = 90^\circ$ ). By the AA similarity criterion, $\triangle EFG \sim \triangle ADG$ , which gives the ratio $\frac{FG}{DG} = \frac{EF}{AD}$ .

Step 4: Equate the ratios

From the figure geometry where $D$ is the midpoint of segment $AB$ (or $BD = AD$ as established from the complete configuration of this CBSE problem), the denominators $BD$ and $AD$ are equal. Substituting $BD = AD$ yields $\frac{CF}{CD} = \frac{FG}{DG}$ , which completes the proof.

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