1. **Problem 1:** Given triangle $ABC$ with point $H$ on $AB$ such that $AH \perp AB$ and $CH \perp AB$. We want to understand the properties of $\triangle ACB$ and the point $H$. Since $AH$ and $CH$ are both perpendicular to $AB$, point $H$ is the foot of the perpendiculars from $A$ and $C$ to $AB$. This implies $H$ lies on $AB$ and is the orthogonal projection of both $A$ and $C$ onto $AB$. 2. **Problem 2:** Given $AB = AC$ (triangle $ABC$ is isosceles), with $BE \perp AC$ and $CF \perp AB$. We want to prove $BE = CF$. Since $AB = AC$, triangle $ABC$ is isosceles with $AB = AC$. The altitudes from $B$ and $C$ to the opposite sides $AC$ and $AB$ respectively are equal in length because of symmetry. Therefore, $BE = CF$. 3. **Problem 3:** Given parallelogram $BFED$ inside triangle $ABC$ with $AB = 12$ cm and $BC = 18$ cm. We want to find $x$. Since $BFED$ is a parallelogram, opposite sides are equal and parallel. Using the given lengths and properties of parallelograms, we can set up equations to solve for $x$. Without additional details, assume $x$ relates to a segment proportional to $AB$ or $BC$. For example, if $x$ is a segment on $AB$ or $BC$, use similarity or parallelogram properties to find $x$. 4. **Problem 4:** Given a triangle with segments marked 8, 4, 6 and variable $x$, with an arrow indicating a relationship. Use the triangle proportionality theorem or segment addition to express $x$ in terms of the given lengths. For example, if $x$ is part of a segment divided into parts 8 and 4, then $x = 8 + 4 = 12$ or use ratios if indicated. 5. **Problem 5:** Circle centered at $A$ with chord $DE$ passing through $C$, segments $AE=4$, $ED=5$, $CD=6$, and variable $x$. Use the chord properties and the power of a point theorem. For example, if $C$ lies on chord $DE$, then $CD + DE = CE$ or use the intersecting chords theorem: $AE \cdot ED = CE \cdot CD$. Solve for $x$ accordingly. 6. **Problem 6:** Triangle $ABC$ with $BC=48$ mm, height $AH=16$ mm, and $QF = \frac{9}{5} QD$. Find $QF$ and $QD$. Given the ratio $QF = \frac{9}{5} QD$, let $QD = t$. Then $QF = \frac{9}{5} t$. The sum $QF + QD = t + \frac{9}{5} t = \frac{14}{5} t$. If this sum corresponds to a segment length related to $BC$ or another known length, solve for $t$ and then find $QF$. **Final answers:** 1) Point $H$ is the foot of perpendiculars from $A$ and $C$ to $AB$. 2) $BE = CF$ because triangle $ABC$ is isosceles with $AB = AC$. 3) $x$ depends on parallelogram properties; more data needed for exact value. 4) $x$ can be found by segment addition or proportionality; more data needed. 5) Use chord and circle theorems to solve for $x$; more data needed. 6) Let $QD = t$, then $QF = \frac{9}{5} t$, solve using segment sums related to $BC$.