Concept:
[tex]\text{Two figures are said to be congruent if they are exactly same; Two figures}\\\text{are said to be similar if they have a same shape.}\\\text{Below, I've shown the differences between congruent and similar triangles in}\\\text{the figure. }k\text{ in the figure is a positive number. }[/tex]
[tex]\text{You can understand the similar figures by imagining two circles, one with}\\\text{radius 2cm and another with 3cm. Or, one square with length 2cm and another}\\\text{with length 6cm. }[/tex]
[tex]\bold{How\ to\ know\ if\ two\ triangles\ are\ similar?}[/tex]
[tex]\text{In the second figure, if }\angle\text{A}=\angle\text{D and }\angle\text{B}=\angle\text{E (or }\angle\text{C}=\angle\text{F}), \text{then the two}\\\text{corresponding angles of the triangles will be equal. Then you may say say that}\\\text{the triangles ABC and FED are similar triangles.}[/tex]
[tex]\text{This is also called }\bold{similar\ triangles\ by\ A.A.\ axiom.}\\\text{In this case, you may say: }\\\\\dfrac{\text{side opposite to }\angle\text{A}}{\text{side opposite to }\angle\text{D}}=\dfrac{\text{side opposite to }\angle\text{B}}{\text{side opposite to }\angle\text{E}}=\dfrac{\text{side opposite to }\angle\text{C}}{\text{side opposite to }\angle\text{F}}[/tex]
[tex](\because\ \angle\text{A}=\angle \text{D})\qquad\quad\quad(\because\ \angle\text{B}=\angle \text{E})\qquad\quad\quad(\because\ \angle\text{C}=\angle \text{F})\\\\\text{or, }\dfrac{\text{CB}}{\text{EF}}=\dfrac{\text{AC}}{\text{FD}}=\dfrac{\text{AB}}{\text{DE}}[/tex]
[tex]\text{This expression is based on the theorem - corresponding sides of similar}\\\text{triangles are proportional to each other.}[/tex]
[tex]\text{Similar figures are denoted by }\sim \text{ symbol. For example from the figure I}\\\text{posted, }\\\triangle\text{ABC}\sim\triangle\text{FED [By A.A. postulate]}[/tex]
Answer with step-by-step explanation:
[tex]1.\ \text{In the triangles LID and CAP, given,}\\\text{i. }\angle\text{D}=\angle\text{P}\ \ \ (A)\text{ [Given]}\\\text{ii. }\angle\text{I}=\angle\text{A}\ \ \ (A)\text{ [Given]}\\\text{iii. }\triangle\text{LID}\sim\triangle\text{CAP [By A.A. postulate]}[/tex]
[tex]\text{2. }\angle\text{L}=\angle\text{C}\ \ \ [\text{Remaining angles of similar triangles.}][/tex]
[tex]\text{3. }\dfrac{\text{side opposite to }\angle\text{L}}{\text{side opposite to }\angle\text{C}}=\dfrac{\text{side opposite to }\angle\text{I}}{\text{side opposite to }\angle\text{A}}\\\\\text{}\quad\text{(You don't need to write this.)}[/tex]
[tex]\dfrac{\text{ID}}{\text{AP}}=\dfrac{\text{LD}}{\text{CP}}\\\\\text{or, }\dfrac{g}{9}=\dfrac{4}{6}=\dfrac{2}{3}\\\\\text{or, }3g=18\\\text{or, }g=6[/tex]
