\( \triangle ABC \) मध्ये, कारण \( BD \parallel \angle ABC \) चा दुसरंयक आहे.
A - D - C, \( \overline{DE} \parallel BC, A - E - B \), तर \(\frac{AB}{BC} = \frac{AE}{EB}\) हे सिद्ध करण्यासाठी खालिल कृत्य पूर्ण करा.
प्रथम, \( \triangle ABC \) मध्ये \( BD \parallel \angle ABC \) आहे. त्यामुळे, त्याचे प्रमाणाचे नियम वापरून, \[ \frac{AD}{DC} = \frac{BC}{DC} \text{(I) } \]
आणि, \[ \triangle ABC \, मध्ये, DE \parallel BC असल्याने, \]
\[ \frac{AD}{EB} = \frac{AB}{EB} \text{(II) } \] आता, प्रमाणाने सिद्ध करत आहोत, \[ \frac{AB}{EB} = \left( \frac{AD}{DC} \right) \text{(I) आणि (II) यांचा उपयोग करा.} \]
In the adjoining figure, \(PQ \parallel XY \parallel BC\), \(AP=2\ \text{cm}, PX=1.5\ \text{cm}, BX=4\ \text{cm}\). If \(QY=0.75\ \text{cm}\), then \(AQ+CY =\)
In the adjoining figure, \( \triangle CAB \) is a right triangle, right angled at A and \( AD \perp BC \). Prove that \( \triangle ADB \sim \triangle CDA \). Further, if \( BC = 10 \text{ cm} \) and \( CD = 2 \text{ cm} \), find the length of } \( AD \).
If a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to the third side. State and prove the converse of the above statement.
In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\). 
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\)) 
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.